What is the length of the radius and the coordinates of the center of the circle defined by the equation $(x+7)^2+(y-3)^2=121$ ?

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Answer 1 · 2017-03-14T12:54:15

The radius is $11 (14-3)$ and the coordinates of the centre is ( $7,3$ )

Opening up the equation,
$(x+7)^2 + (y-3)^2 = 121$
$x^2+14x+49+y^2-6y+9 = 121$
$y^2-6y = 63-x^2+14x$

Find the x-intercepts, and the midpoint to find x-line of symmetry,
When $y = 0$ ,
$x^2-14x-63 = 0$
$x=17.58300524 or x=-3.58300524$

$(17.58300524-3.58300524)/2 = 7$

Find the highest and lowest point and midpoint,
When $x =7$ ,
$y^2-6y-112=0$
$y = 14 or y = -8$

$(14-8)/2 = 3$

Hence, the radius is $11 (14-3)$ and the coordinates of the centre is ( $7,3$ )

What is the length of the radius and the coordinates of the center of the circle defined by the equation (x+7)^2+(y-3)^2=121(x+7)^2+(y-3)^2=121?